A) \[-\frac{7}{9}\]
B) \[-\frac{9}{7}\]
C) \[\frac{7}{9}\]
D) \[\frac{9}{7}\]
Correct Answer: D
Solution :
\[{{T}_{r+1}}={}^{\text{9}}{{C}_{r}}{{(3)}^{9-r}}{{(ax)}^{r}}={}^{\text{9}}{{C}_{r}}{{(3)}^{9-r}}{{a}^{r}}{{x}^{r}}\] \[\therefore \] Coefficient of \[{{x}^{r}}\]= \[{}^{9}{{C}_{r}}{{3}^{9-r}}{{a}^{r}}\] Hence, coefficient of \[{{x}^{2}}={}^{9}{{C}_{2}}{{3}^{9-2}}{{a}^{2}}\] and coefficient of \[{{x}^{\text{3}}}\] = \[{}^{9}{{C}_{3}}{{3}^{9-3}}{{a}^{3}}\] So, we must have \[{}^{9}{{C}_{2}}{{3}^{7}}{{a}^{2}}={}^{9}{{C}_{3}}{{3}^{6}}{{a}^{3}}\] Þ \[\frac{9.8}{1.2}.3=\frac{9.8.7}{1.2.3}.a\,\,\,\Rightarrow \,\,a=\frac{9}{7}.\]
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